A uniform horizontal rod of mass M and length l rotates with angular velocity w (omega) about a vertical axis through its center. Attached to each end of the rod is a small mass m. Determine the angular momentum of the system about the axis. Write you're answer in terms of M, m, l and w (omega).
College Physics - drwls, Sunday, October 21, 2012 at 1:50am
The moment of inertia of the rod with the two masses at each end is:
I = M*L^2/12 + 2*m*(L/2)^2
= M*L^2/12 + m*L^2/2
= (M + 6m)*L^2/12
The angular momentum is
I*w = (M+6m)*w*L^2/12.
College Physics - ann, Monday, October 29, 2012 at 10:48pm
If Wire A has a poterntial difference of 50V across it and carries a current of 2A. Wire B has a potential difference of 100V across it and also carries a current of 2A. Compare the resistances, rate of flow of charge, and rate of flow of energy in the two wires?